Question

分解因式(x+1)(x+2)(x+3)(x+4)-120

用换元法怎么做？

Answer 1

(x+1)(x+2)(x+3)(x+4)-120
=[(x+1)(x+4)][(x+2)(x+3)]-120
=(x²+5x+4)(x²+5x+6)-120
令y=x²+5x，则
原式=(y+4)(y+6)-120
=y²+10y-96
=(y+16)(y-6)
=(x²+5x+16)(x²+5x-6)
=(x²+5x+16)(x+6)(x-1)

Answer 2

(x-1)(x-2)(x-3)(x-4)-120
=(x-1)(x-4)(x-2)(x-3)-120(首位相乘,使有公共项x^2-5x)
=(x^2-5x+4)(x^2-5x+6)-120
=(x^2-5x)^2+(4+6)(x^2-5x)-120+24(把x^2-5x看作一项,换元)
=(x^2-5x)^2+10(x^2-5x)-96(再使用十字相乘)
=(x^2-5x+16)(x^2-5x-6)(同上)
=(x^2-5x+16)(x-6)(x+1)

Answer 3

(x+1)(x+2)(x+3)(x+4)-120
=[(x+1)(x+4)]*[(x+2)(x+3)]-120
=(x^2+5x+4)(x^2+5x+6)-120
=[(x^2+5x)+4][(x^2+5x)+6]-120

=(x^2+5x)^2+10(x^2+5x)+24-120
=(x^2+5x)^2+10(x^2+5x)-96
=(x^2+5x+16)(x^2+5x-6)
=(x^2+5x+16)(x+6)(x-1)